Query Processing. Query Processing Components

Transcription

1 Query Processing high level user query (SQL) Query Compiler Query Processor Plan Plan Cost Generator Estimator low level data manipulation commands (execution plan) Plan Evaluator 7-1 Query Processing Components Query language that is used SQL: intergalactic dataspeak Query execution methodology The steps that one goes through in executing high-level (declarative) user queries. Query optimization How do we determine a good execution plan? 7-2

2 What are we trying to do? Consider query For each project whose budget is greater than $ and which employs more than two employees, list the names and titles of employees. In SQL SELECT Ename, Title FROM Emp, Project, Works WHERE Budget > AND Emp.Eno=Works.Eno AND Project.Pno=Works.Pno AND Project.Pno IN (SELECT w.pno FROM Works w GROUP BY w.pno HAVING SUM(*) > 2) How to execute this query? 7-3 A Possible Execution Plan 1. T 1 Scan Project table and select all tuples with Budget value > T 2 Join T 1 with the Works relation 3. T 3 Join T 2 with the Emp relation 4. T 4 Group tuples of T 3 over Pno 5. Scan tuples in each group of T 4 and for groups that have more than 2 tuples, Project over Ename, Title Note: Overly simplified we ll detail later. 7-4

3 Pictorial Representation Π Ename, Title T 1 σ Budget> Project T 2 T 4 Group by T 3 Works Emp 1. How do we get this plan? 2. How do we execute each of the nodes? Π Ename, Title (Group Pno,Eno (Emp(σ Budget> ProjectWorks))) 7-5 Query Processing Methodology SQL Queries Normalization Normalization Analysis Analysis Simplification Simplification System Catalog Restructuring Restructuring Optimization Optimization Optimal Execution Plan 7-6

4 Query Normalization Lexical and syntactic analysis check validity (similar to compilers) check for attributes and relations type checking on the qualification Put into (query normal form Conjunctive normal form (p 11 p 12 p 1n ) (p m1 p m2 p mn ) Disjunctive normal form (p 11 p 12 p 1n ) (p m1 p m2 p mn ) OR's mapped into union AND's mapped into join or selection 7-7 Analysis Refute incorrect queries Type incorrect If any of its attribute or relation names are not defined in the global schema If operations are applied to attributes of the wrong type Semantically incorrect Components do not contribute in any way to the generation of the result Only a subset of relational calculus queries can be tested for correctness Those that do not contain disjunction and negation To detect connection graph (query graph) join graph 7-8

5 Emp.Eno=Works.Eno Title = Programmer Ename Analysis Example SELECT Ename,Resp FROM Emp, Works, Project WHERE Emp.Eno = Works.Eno AND Works.Pno = Project.Pno AND Pname = CAD/CAM AND Dur > 36 AND Title = Programmer Query graph Works Resp RESULT Dur>36 Works.Pno=Project.Pno Emp.Eno=Works.Eno Pname= CAD/CAM Join graph Works Works.Pno=Project.Pno Emp Project Emp Project 7-9 Analysis If the query graph is not connected, the query may be wrong. SELECT Ename,Resp FROM Emp, Works, Project WHERE Emp.Eno = Works.Eno AND Pname = CAD/CAM AND Dur > 36 AND Title = Programmer Works Emp Resp Project Ename RESULT Pname= CAD/CAM 7-10

6 Why simplify? The simpler the query, the easier (and more efficient) it is to execute it How? Use transformation rules elimination of redundancy idempotency rules p 1 ( p 1 ) false p 1 (p 1 p 2 ) p 1 Simplification p 1 false p 1 application of transitivity use of integrity rules 7-11 Simplification Example SELECT FROM WHERE OR AND OR AND SELECT FROM WHERE Title Emp Ename = J. Doe (NOT(Title = Programmer ) (Title = Programmer Title = Elect. Eng. ) NOT(Title = Elect. Eng. )) Title Emp Ename = J. Doe 7-12

7 Convert SQL to relational algebra Make use of query trees Example Restructuring SELECT Ename FROM Emp, Works, Project WHERE Emp.Eno = Works.Eno AND Works.Pno = Project.Pno AND Ename <> J. Doe AND Pname = CAD/CAM AND (Dur = 12 OR Dur = 24) Π ENAME σ DUR=12 OR DUR=24 σ PNAME= CAD/CAM σ ENAME J. DOE PNO ENO Project Select Join Project Works Emp 7-13 How to implement operators Selection (assume R has n pages) Scan without an index O(n) Scan with index B + index O(logn) Hash index O(1) Projection Without duplicate elimination O(n) With duplicate elimination Sorting-based O(nlogn) Hash-based O(n+t) where t is the result of hashing phase 7-14

8 Join How to implement operators (cont d) Nested loop join: RS foreach tuple r R do foreach tuple s S do if r==s then add <r,s> to result O(n*m) Improvements possible by page-oriented nested loop join block-oriented nested loop join 7-15 Join How to implement operators (cont d) Index nested loop join: RS foreach tuple r R do use index on join attr. to find tuples of S foreach such tuple s S do add <r,s> to result Sort-merge join Sort R and S on the join attribute Merge the sorted relations Hash join Hash R and S using a common hash function Within each bucket, find tuples where r=s 7-16

9 Index Selection Guidelines Hash vs tree index Hash index on inner is very good for Index Nested Loops. Should be clustered if join column is not key for inner, and inner tuples need to be retrieved. Clustered B+ tree on join column(s) good for Sort- Merge Example 1 SELECT e.ename, w.dur FROM Emp e, Works w WHERE w.resp= Mgr AND e.eno=w.eno Hash index on w.resp supports Mgr selection. Hash index on w.eno allows us to get matching (inner) Emp tuples for each selected (outer) Works tuple. What if WHERE included: AND e.title=`programmer? Could retrieve Emp tuples using index on e.title, then join with Works tuples satisfying Resp selection. 7-18

10 Example 2 SELECT e.ename, w.resp FROM Emp e, Works w WHERE e.age BETWEEN 45 AND 60 AND e.title= Programmer AND e.eno=w.eno Clearly, Emp should be the outer relation. Suggests that we build a hash index on w.eno. What index should we build on Emp? B+ tree on e.age could be used, OR an index on e.title could be used. Only one of these is needed, and which is better depends upon the selectivity of the conditions. As a rule of thumb, equality selections more selective than range selections. As both examples indicate, our choice of indexes is guided by the plan(s) that we expect an optimizer to consider for a query. Have to understand optimizers! 7-19 Examples of Clustering SELECT e.title FROM Emp e WHERE e.age > 40 B+ tree index on e.age can be used to get qualifying tuples. How selective is the condition? Is the index clustered? 7-20

11 Clustering and Joins SELECT e.ename, p.pname FROM Emp e, Project p WHERE p.budget= AND e.city=p.city Clustering is especially important when accessing inner tuples in Index Nested Loop join. Should make index on e.city clustered. Suppose that the WHERE clause is instead: WHERE e.title= Programmer AND e.city=p.city If many employees are Programmers, Sort-Merge join may be worth considering. A clustered index on p.city would help. Summary: Clustering is useful whenever many tuples are to be retrieved Selecting Alternatives SELECT Ename FROM Emp e,works w WHERE e.eno = w.eno AND w.dur > 37 Strategy 1 Π ENAME (σ DUR>37 EMP.ENO=ASG.ENO (Emp Works)) Strategy 2 Π ENAME (Emp ENO (σ DUR>37 (Works))) Strategy 2 is better because It avoids Cartesian product It selects a subset of Works before joining How to determine the better alternative? 7-22

12 Query Optimization Issues Types of Optimizers Exhaustive search cost-based optimal combinatorial complexity in the number of relations Heuristics not optimal regroup common sub-expressions perform selection, projection as early as possible reorder operations to reduce intermediate relation size optimize individual operations 7-23 Query Optimization Issues Optimization Granularity Single query at a time cannot use common intermediate results Multiple queries at a time efficient if many similar queries decision space is much larger 7-24

13 Query Optimization Issues Optimization Timing Static compilation optimize prior to the execution difficult to estimate the size of the intermediate results error propagation can amortize over many executions Dynamic run time optimization exact information on the intermediate relation sizes have to reoptimize for multiple executions Hybrid compile using a static algorithm if the error in estimate sizes > threshold, reoptimize at run time 7-25 Query Optimization Issues Statistics Relation cardinality size of a tuple fraction of tuples participating in a join with another relation Attribute cardinality of domain actual number of distinct values Common assumptions independence between different attribute values uniform distribution of attribute values within their domain 7-26

14 Query Optimization Components Cost function (in terms of time) I/O cost + CPU cost These might have different weights Can also maximize throughput Solution space The set of equivalent algebra expressions (query trees). Search algorithm How do we move inside the solution space? Exhaustive search, heuristic algorithms (iterative improvement, simulated annealing, genetic, ) 7-27 Cost Calculation Cost function takes CPU and I/O processing into account Instruction and I/O path lengths Estimate the cost of executing each node of the query tree Is pipelining used or are temporary relations created? Estimate the size of the result of each node Selectivity of operations reduction factor Error propagation is possible 7-28

15 Intermediate Relation Sizes Selection size(r) = card(r) length(r) card(σ F (R)) = SF σ (F) card(r) where 1 S F σ (A = value) = card( A (R)) max(a) value S F σ (A > value) = max(a) min(a) value min(a) S F σ (A < value) = max(a) min(a) SF σ (p(a i ) p(a j )) = SF σ (p(a i )) SF σ (p(a j )) SF σ (p(a i ) p(a j )) = SF σ (p(a i )) + SF σ (p(a j )) (SF σ (p(a i )) SF σ (p(a j ))) SF σ (A value) = SF σ (A= value) card({values}) 7-29 Intermediate Relation Sizes Projection card(π A (R))=card(R) Cartesian Product Union card(r S) = card(r) card(s) upper bound: card(r S) = card(r) + card(s) lower bound: card(r S) = max{card(r), card(s)} Set Difference upper bound: card(r S) = card(r) lower bound:

16 Intermediate Relation Size Join Special case: A is a key of R and B is a foreign key of S; card(r A=B S) = card(s) More general: card(r S) = SF J card(r) card(s) 7-31 Search Space Characterized by equivalent query plans Equivalence is defined in terms of equivalent query results Equivalent plans are generated by means of algebraic transformation rules The cost of each plan may be different Focus on joins 7-32

17 Search Space Join Trees For N relations, there are O(N!) equivalent join trees that can be obtained by applying commutativity and associativity rules SELECT FROM WHERE AND Emp Ename,Resp Emp, Works, Project Project Emp.Eno=Works.Eno Works.PNO=Project.PNO Works Works Project Emp Works Project Emp 7-33 Transformation Rules Commutativity of binary operations R S S R R S S R R S S R Associativity of binary operations ( R S ) T R (S T) ( R S ) T R (S T ) Idempotence of unary operations Π A (Π A (R)) Π A (R) σ p1 (A 1 ) (σ p 2 (A 2 ) (R)) = σ p 1 (A 1 ) p 2 (A 2 ) (R) where R[A] and A' A, A" A and A' A" 7-34

18 Transformation Rules Commuting selection with projection Commuting selection with binary operations σ p(a) (R S) (σ p(a) (R)) S σ p(ai ) (R (A j,b k ) S) (σ p(a i ) (R)) (A j,b k ) S σ p(ai ) (R T) σ p(a i ) (R) σ p(a i ) (T) where A i belongs to R and T Commuting projection with binary operations Π C (R S) Π A (R) Π B (S) Π C (R (Aj,B k ) S) Π A (R) (A j,b k ) Π B (S) Π C (R S) Π C (R) Π C (S) where R[A] and S[B]; C = A' B' where A' A, B' B, A j A', B k B' 7-35 Example Consider the query: Find the names of employees other than J. Doe who worked on the CAD/CAM project for either one or two years. Π ENAME σ DUR=12 OR DUR=24 Project SELECT Ename FROM Project p, Works w, Emp e WHERE w.eno=e.eno AND w.pno=p.pno AND Ename<>`J. Doe AND p.pname=`cad/cam AND (Dur=12 OR Dur=24) σ PNAME= CAD/CAM σ ENAME J. DOE Select Join Project Works Emp 7-36

19 Equivalent Query Π Ename σ Pname=`CAD/CAM (Dur=12 Dur=24) Ename<>`J. DOE Works Project Emp 7-37 Another Equivalent Query Π Ename Π Pno,Ename Π Pno Π Pno,Eno Π Eno,Ename σ Pname = `CAD/CAM σ Dur =12 Dur=24 σ Ename <> `J. Doe Project Works Emp 7-38

20 Search Strategy How to move in the search space. Deterministic Start from base relations and build plans by adding one relation at each step Dynamic programming: breadth-first Greedy: depth-first Randomized Search for optimalities around a particular starting point Trade optimization time for execution time Better when > 5-6 relations Simulated annealing Iterative improvement 7-39 Search Algorithms Restrict the search space Use heuristics E.g., Perform unary operations before binary operations Restrict the shape of the join tree Consider only linear trees, ignore bushy ones Linear Join Tree R 4 Bushy Join Tree R 3 R 1 R 2 R 1 R 2 R 3 R

21 Deterministic Search Strategies R 4 R 3 R 3 R 1 R 2 R 1 R 2 Randomized R 1 R 2 R 3 R 2 R 1 R 2 R 1 R Summary Declarative SQL queries need to be converted into low level execution plans These plans need to be optimized to find the best plan Optimization involves Search space: identifies the alternative plans and alternative execution algorithms for algebra operators This is done by means of transformation rules Cost function: calculates the cost of executing each plan CPU and I/O costs Search algorithm: controls which alternative plans are investigated 7-42